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Question
In a series LCR circuit connected to an a.c. source of voltage v = vmsinωt, use phasor diagram to derive an expression for the current in the circuit. Hence, obtain the expression for the power dissipated in the circuit. Show that power dissipated at resonance is maximum
Solution
The phasor diagram for the series LCR circuit is
The phasor relation for the voltages is
VL + VR + VC = V
Because VC and VL are always along the same line and in opposite directions, they can be combined into a single phasor (VC + VL) which has a magnitude |vCm - vLm|.
Because V is represented as the hypotenuse of a right triangle whose sides are VR and (VC + VL), the Pythagorean theorem gives
`v_m^2=v_(Rm)^2+(V_(Cm)-V_(Lm))^2`
`:.v_m^2=(i_mR)^2+(i_mX_c-i_mX_L)^2`
`:.v_m^2=i_m^2[R^2+(X_C-X_L)^2]`
`:.i_m=v_m/sqrt(R^2+(X_C-X_L)^2)`
This is the expression for the current in the circuit.
The instantaneous power in the circuit is
P = vi = (vmsinωt)(imsin(ωt+Φ))
`:. P=(v_mi_m)/2(sinomegat)(sin(omegat+phi))`
`:.P=(v_mi_m)/2(cosphi-cos(2omegat+phi))`
The average power over a cycle is given by the average of the two terms in the RHS of the above equation.
It is only the second term (cos(2ωt+)) which is time-dependent. Its average is zero. Therefore, we get
`P=(v_mi_m)/2cosphi=v_m/sqrt2xxi_m/sqrt2cosphi`
∴ P = VI cosΦ
It can also be written as P = I2ZcosΦ
At resonance XC − XL = 0 and Φ = 0. Therefore, cosΦ = 1 and power is P=I2Z=I2R.
That is, maximum power is dissipated in a circuit at resonance.
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